Problem: John Doe has an n × m table. John Doe can paint points in some table cells, not more than one point in one table cell. John Doe wants to use such operations to make each square subtable of size n × n have exactly k points.

John Doe wondered, how many distinct ways to fill the table with points are there, provided that the condition must hold. As this number can be rather large, John Doe asks to find its remainder after dividing by 1000000007 (109 + 7).

You should assume that John always paints a point exactly in the center of some cell. Two ways to fill a table are considered distinct, if there exists a table cell, that has a point in one way and doesn't have it in the other.

Input

A single line contains space-separated integers n, m, k (1 ≤ n ≤ 100; n ≤ m ≤ 1018; 0 ≤ k ≤ n2) — the number of rows of the table, the number of columns of the table and the number of points each square must contain.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. 

Output

In a single line print a single integer — the remainder from dividing the described number of ways by 1000000007 (109 + 7).

Examples

Input

5 6 1


Output

45

Note

Let's consider the first test case: 

<image> The gray area belongs to both 5 × 5 squares. So, if it has one point, then there shouldn't be points in any other place. If one of the white areas has a point, then the other one also must have a point. Thus, there are about 20 variants, where the point lies in the gray area and 25 variants, where each of the white areas contains a point. Overall there are 45 variants.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class TableCF232B {
    
    
    
    /************************ SOLUTION STARTS HERE ************************/
    
    static final int mod = (int) (1e9) + 7; // Default
    static class MM {       // MM (Modular Math) class 
        static int sub(int a, int b) {return (a - b  + mod) % mod;}
        static int mul(int a, int b) {return (int) ((1L * a * b ) % mod);}
        static int add(int a, int b) {return (a + b) % mod;}
        static int div(int a, int b) {return mul(a, modInverse(b));}
        static int modInverse(int n)  {return modPow(n, mod - 2);} // Fermat's little theorem
        static int modPow(long a , long b) {
            long pow = 1;
            while(b > 0) {
                if((b & 1L) == 1)
                    pow = ((pow * a) % mod);

                a = ((a * a) % (mod));
                b >>= 1;
            }
            return (int) pow;
        }
    }
    
    static final int MAX = 128;                    
    static int fact[] = new int[MAX];
    static int invFact[] = new int[MAX];
    static {
        fact[0] = fact[1] = 1;
        for (int i = 2; i < MAX; i++)
            fact[i] = MM.mul(i, fact[i - 1]);
        for (int i = 0; i < MAX; i++)
            invFact[i] = MM.modInverse(fact[i]);
    }

    static int nCr(int n, int r) { // Precompute inv Factorials (Dont compute every time) 
        return MM.mul(fact[n], MM.mul(invFact[r], invFact[n - r]));
    }
    
    static long m;
    static int n;
    static int memo[][];
    static int rec(int idx , int k) {
        if(idx == n)
            return k == 0 ? 1 : 0;
        else if(memo[idx][k] != -1)
            return memo[idx][k];
        else {
            long pow = (m / n) + (idx < m % n ? 1 : 0);
            int sum = 0;
            for(int i = 0; i <= Math.min(k , n); i++) 
                sum = MM.add(sum, MM.mul(rec(idx + 1, k - i), MM.modPow(nCr(n, i), pow)));
            
            return memo[idx][k] = sum;
        }
    }
    
    private static void solve() {
        
        n = nextInt();
        m = nextLong();
        int k = nextInt();
        
        memo = new int[n][k + 1];
        for(int temp[] : memo) Arrays.fill(temp, -1);
        
        long st = System.nanoTime();
        println(rec(0, k));
        System.out.println("Time : " + (System.nanoTime() - st) / 1e9);
    }

    private static void solve2() {
        
        int n = nextInt();
        long m = nextLong();
        int k = nextInt();
        long quo = m / n;
        int rem = (int) (m % n);
        int cache[][] = new int[Math.min(n , k) + 1][2];
        for(int i = 0; i <= Math.min(n , k); i++) {
            int comb = nCr(n, i);
            cache[i][0] = MM.modPow(comb , quo);
            cache[i][1] = MM.mul(cache[i][0], comb);
        }
        int DP[] = new int[k + 1];
        DP[0] = 1;   // base case
        for(int i = 1; i <= Math.min(n , k); i++) 
            DP[i] = cache[i][0];
        
        for(int i = n - 2; i >= 0; i--) {
            int temp[] = new int[k + 1];
            int flag  = (i < rem ? 1 : 0);
            for(int kk = 0; kk <= k; kk++)
                for(int j = 0; j <= Math.min(n , kk); j++)
                    temp[kk] = MM.add(temp[kk], MM.mul(cache[j][flag], DP[kk - j]));
            DP = temp;
        }
        
        println(DP[k]);
    }
    
    
    /************************ SOLUTION ENDS HERE ************************/
    
    
    
    
    
    /************************ TEMPLATE STARTS HERE **********************/
    
    public static void main(String[] args) throws IOException {
        reader = new BufferedReader(new InputStreamReader(System.in));
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
        st     = null;
        solve2();
        reader.close();
        writer.close();
    }
    
    static BufferedReader reader;
    static PrintWriter    writer;
    static StringTokenizer st;
    
    static String next()
    {while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}            
    st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
    static String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}             
    static int    nextInt()   {return Integer.parseInt(next());}
    static long   nextLong()  {return Long.parseLong(next());}     
    static double nextDouble(){return Double.parseDouble(next());}
    static char   nextChar()  {return next().charAt(0);}
    static int[]  nextIntArray(int n)         {int[] a= new int[n];   int i=0;while(i<n){a[i++]=nextInt();}  return a;}
    static long[] nextLongArray(int n)        {long[]a= new long[n];  int i=0;while(i<n){a[i++]=nextLong();} return a;}    
    static int[]  nextIntArrayOneBased(int n) {int[] a= new int[n+1]; int i=1;while(i<=n){a[i++]=nextInt();} return a;}            
    static long[] nextLongArrayOneBased(int n){long[]a= new long[n+1];int i=1;while(i<=n){a[i++]=nextLong();}return a;}            
    static void   print(Object o)  { writer.print(o);  }
    static void   println(Object o){ writer.println(o);}
    
    /************************ TEMPLATE ENDS HERE ************************/
    
}